Freiling G., Yurko V. Lectures on Differential Equations of Mathematical Physics: A First Course

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212 G.

Freiling and V. Yurko

problem ( A

The reason is that we seek a solution of problem ( A

) in the form (4.6.24),

but by Theorem 4.6.1,

F(x) = O

kxk

, x →∞.

Thus,

do not take into account solutions of the class O

kxk

. Therefore,

will act

as follows.

Fix x

∈ D. Let us seek a solution of the problem ( A

) in the form

u(x) = F(x) +

kx −x

where F(x) has

the

form (4.6.24), α = const. Clearly, u(x) is harmonic in D

, u(∞) = 0

and u(x) ∈C(

). Therefore,

for

the function u(x) to be a solution of ( A

), it is necessary

that u

|Σ

= ϕ

, i.e.

(x) = F

(x) +

kx −x

, x ∈Σ.

Since F

(x)

= F(x)

+ 2π f (x), we arrive at the equation:

f (x) =

(x)

2π

−

2πkx −x

−

K(x,ξ) f (ξ)d

s. (

4.6.28)

According to Theorem 4.6.12, equation (4.6.28) is solvable if and only if

(ξ)

2π

−

2πkξ −x

(ξ)d

s = 0

Since x

∈ D and q

is the Roben potential, we have

(ξ)

kξ −x

s = 1

and consequently,

α =

(ξ)q

(ξ)ds.

Thus, we have proved the following assertion.

Theorem 4.6.14. For any continuous function ϕ

(x) the solution of the exterior

Dirichlet problem ( A

) exists, is unique and has the form

u(x) = F(x) +

kx −x

(ξ)q

(ξ)d

, x ∈D

where q

(x) is the Roben potential, F(x) has the form (4.6.24) , and f (x) is a solution

of the integral equation (4.6.28) .

We note that equation (4.6.28) has an inﬁnite number of solutions but F(x) is unique.

Indeed, if

f (x) is another solution (different from f (x) ) of equation (4.6.28), then

f (x) −

f (x) ≡C, and by virtue of Lemma 4.6.7,

f (ξ) −

f (ξ)

∂

(

)

∂n

s = C

∂

(

)

∂n

s = 0

, x ∈D

Elliptic P

artial Differential Equations 213

4.7. The

Variational Method

1. The variational Dirichlet principle

Let D ⊂ R

be a bounded domain with the boundary Σ ∈PC

. We consider the Dirichlet

problem

∆u = 0 (x ∈ D),

|Σ

= ϕ(x), ϕ(x) ∈C(Σ).

)

(4.7.1)

Denote

:= {u : u ∈C(

D), u ∈C

(D), u

|Σ

= ϕ}.

function u

(x) is called a solution of problem (4.7.1) if u ∈ R

and ∆u = 0 in D. We

consider the Dirichlet functional

Φ(u) :=

∑

k=1

∂u

∂x

x (4

.7.2)

and formulate the variational extremal problem

Φ(u) → inf, u ∈ R

. (4.7.3)

Clearly, Φ(u) ≥ 0, and consequently, there exists inf

u∈R

Φ(u). Suppose that there exists a

function u

∗

∈ R

such that

Φ(u

∗

) = inf

u∈R

Φ(u).

Then u

∗

satisﬁes the Euler equation. For the functional (4.7.2) the Euler equation has the

form ∆u = 0. Therefore, if u

∗

is a solution of the variational problem (4.7.3), then u

∗

a solution of problem (4.7.1). Thus, we replace (4.7.1) by the variational problem (4.7.3).

This approach is fruitful since one can use methods of the variational calculus.

Criticism. 1) It can happen that infΦ(u) = ∞, i.e. Φ(u) = ∞ ∀u ∈R

. An example of

such a function is due to Hadamard (see [8, Chapter 22]). In this case the variational method

does not work (although a solution of the problem (4.7.1) can exist as in the Hadamard

example). In order to apply the variational method we impose an additional condition.

Denote Q

:= {u ∈R

: Φ(u) < ∞}.

Continuability (extendability) condition: Q

0. (4.7.4)

For such ϕ we have inf Φ(u) < ∞. In particular, if Σ is smooth, and ϕ ∈C

, then condi-

tion (4.7.4) is fulﬁlled automatically.

2) We suppose that problem (4.7.3) has a solution. This is not always valid, i.e. the

inﬁmum in (4.7.3) is not always attained (see [8, Chapter 22]). However, if one takes simply

connected domains and extends the notion of the solution, then the variational method gives

us the existence and the uniqueness of the solution of the Dirichlet problem, and also a

constructive procedure for its solution.

214 G.

Freiling and V. Yurko

2. Operator

equations in a Hilbert space

Let H be a real Hilbert space with the scalar product (·, ·) and with the norm k·k. Let

A : D

→ E

be a symmetric operator in H with an everywhere dense domain D

Deﬁnition 4.7.1. The operator A is called positive deﬁnite ( A > 0 ), if (Au,u) ≥γkuk

for some γ > 0 and for all u ∈D

Everywhere below we assume that A > 0. In particular, this yields that on E

there

exists the inverse operator A

−1

. We consider in H the equation

Au = f , f ∈ H. (4.7.5)

An element u

∗

is called a solution of (4.7.5), if u

∗

∈ D

and Au

∗

= f . The following

theorem is obvious.

Theorem 4.7.1. 1) If a solution of (4.7.5) exists, then it is unique.

2) A solution of (4.7.5) exists if and only if f ∈ E

. In particular, if E

= H, then a

solution of (4.7.5) exists for all f ∈ H.

Consider the functional

F(u) = (Au,u) −2(u, f ), u ∈D

(4.7.6)

on D

and the extremal problem

F(u) → inf, u ∈ D

. (4.7.7)

An element u

∗

∈ D

is called a solution of the problem (4.7.7), if F(u

∗

) = inf

u∈D

F(u).

Theorem 4.7.2. The problems (4.7.5) and (4.7.7) are equivalent, i.e. they are solvable

or unsolvable simultaneously, and u

∗

is a solution of (4.7.5) if and only if u

∗

is a solution

of (4.7.7).

Proof. 1) Let Au

∗

= f , u

∗

∈ D

. Let v ∈D

, η = v −u

∗

, i.e. v = u

∗

+ η. Then

F(v) = (A(u

∗

+ η),u

∗

+ η) −2(u

∗

+ η, f ) = F(u

∗

) + (Aη,η) > F(u

∗

and consequently, u

∗

is a solution of (4.7.7).

2) Let u

∗

be a solution of (4.7.7), and let η ∈D

, λ = const. . Then u

∗

+λη ∈D

, and

consequently, F(u

∗

+ λη) ≥ F(u

∗

). This yields 2λ(Au

∗

− f ,η) + λ

(Aη,η) ≥ 0. This is

possible only if (Au

∗

− f , η) = 0 for all η ∈D

. Since D

is everywhere dense, we have

∗

= f , and Theorem 4.7.2 is proved. 2

Remark 4.7.1. If E

6= H, f /∈ E

, then the problems (4.7.5) and (4.7.7) have no

solutions. We extend the notion of the solution. For this purpose we introduce the so-called

energy space.

We introduce a new scalar product and the corresponding norm: [u, v] := (Au,v) , |u|

[u,u] , u,v ∈ D

. Then we complete this space by taking its closure, i.e. by including the

limit elements with respect to its norm. As a result we obtain a new Hilbert space which is

denoted by H

and is called the energy space. It can be shown [9] that H

⊂ H.

Elliptic P

artial Differential Equations 215

Example 4.7.1. Let H = L

,1) ,

Au = −u

(x), u(0) = u(1) = 0, D

= {u ∈C

[0,1] : u(0) = u(1) = 0}.

Then

(Au,u) =

(x)

dx.

Since

u(x) =

(x)dx,

we have |u(x)| ≤ ku

, and consequently, kuk

≤ ku

, i.e. A > 0. Moreover,

= {u(x) ∈ AC[0,1] : u

(x) ∈ L

(0,1), u(0) = u(1) = 0}.

We extend the domain of the functional F(u) of the form (4.7.6) to H

, namely, we

consider the functional

F(u) = [u,u] −2(u, f ), u ∈H

Since

|u|

= (Au,u) ≥ γkuk

we have

|(u, f )| ≤ kf k·kuk ≤ γ

−1/2

kf k·|u|,

and consequently, the functional (u, f ) is bounded in H

. By the Riesz representation

theorem, there exists a unique element f

∗

∈ H

such that (u, f ) = [u, f

∗

]. Therefore the

functional takes the form

F(u) = [u,u] −2[u, f

∗

], u ∈H

. (4.7.8)

Consider the extremum problem

F(u) → inf, u ∈ H

. (4.7.9)

An element u

∗

∈ H

is called a solution of (4.7.9), if F(u

∗

) = inf

u∈H

F(u).

Theorem 4.7.3. For each f

∗

∈ H

problem (4.7.9) has a unique solution u

∗

, and

∗

= f

∗

Indeed, by virtue of (4.7.8),

F(u) = [u − f

∗

,u − f

∗

] −[ f

∗

, f

∗

] = |u − f

∗

−|f

∗

This yields the assertion of the theorem.

Deﬁnition 4.7.2. An element u

∗

∈H

, which gives a solution of (4.7.9) (i.e. u

∗

= f

∗

is called a generalized solution of problem (4.7.5).

One can give another (equivalent) deﬁnition of a generalized solution of problem

(4.7.5). Indeed, from (4.7.5) we infer (Au,v) = ( f ,v) , v ∈ H

or [u,v] = ( f ,v).

216 G.

Freiling and V. Yurko

Deﬁnition 4.7.3. An

element u

∗

∈ H

is called a generalized solution of problem

(4.7.5) if [u

∗

,v] = ( f ,v) for all v ∈ H

These deﬁnitions are equivalent since ( f ,v) = [ f

∗

,v].

Theorem 4.7.4. For each f ∈ H a generalized solution of problem (4.7.5) exists, is

unique and has the form u

∗

= f

∗

3. Solution of the Dirichlet problem

Next we apply the theory developed above for the solution of the Dirichlet problem. We

conﬁne ourselves to the cases n = 2 and n = 3. Let us consider the following Dirichlet

problem:

−∆u = f (x ∈ D), f ∈L

(D),

|Σ

= 0,

)

(4.7.10)

where D is a bounded simply connected domain. Denote

R := {u : u ∈W

(D), u

|Σ

= 0} =: W

2,0

By the embedding theorem u ∈C(

D).

Deﬁnition

4.7.4. A

function u

∗

is called a(classical) solution of problem (4.7.10), if

∗

∈ R and −∆u

∗

= f .

Problem (4.7.10) is a particular case of problem (4.7.5), where H = L

(D), A =

−∆, D

= R. It is known [9] that A > 0. In this case

[u,v] =

∑

∂u

∂x

∂v

∂x

One

can show that H

= W

1,0

Deﬁnition 4.7.5. A function u

∗

is called a generalized solution of problem (4.7.10), if

∗

∈W

1,0

and [u

∗

,v] = ( f ,v) for all v ∈W

1,0

By virtue of Theorem 4.7.4, for any f ∈ L

(D) a generalized solution of problem

(4.7.10) exists and is unique. We note that for (4.7.10) a generalized solution is also a

classical one. In other words, the following assertion is valid [9]:

Theorem 4.7.5. Let u

∗

∈W

1,0

be a generalized solution of (4.7.10) . Then u

∗

∈W

2,0

and −∆u

∗

= f , i.e. u

∗

is a classical solution of (4.7.10) .

Corollary 4.7.1. For any f ∈ L

(D), a classical solution of problem (4.7.10) exists

and is unique.

Now we consider the Dirichlet problem for the homogeneous equation:

∆u = 0 (x ∈ D),

|Σ

= ϕ(x), ϕ(x) ∈C(Σ).

)

(4.7.11)

Elliptic P

artial Differential Equations 217

Denote R

:= {u : u ∈ W

(D), u

|Σ

= ϕ}. If u ∈ R

, then by

the embedding theorem

u ∈C(

D).

Deﬁnition

4.7.6. A

function u

∗

is called a solution of problem (4.7.11) if u

∗

∈ R

and ∆u

∗

= 0 in D.

Denote Q

:= {u ∈ R

: Φ(u) < ∞}, where Φ(u) is deﬁned by (4.7.2). Let Q

i.e. the continuability (extendability) condition (4.7.4) is fulﬁlled. Fix w ∈ Q

. Clearly, if

u ∈ R

, then v := u −w ∈ R, and conversely, if v ∈R, then u := v + w ∈R

. This yields

that the function u

∗

is a solution of problem (4.7.11) if and only if the function v

∗

= u

∗

−w

is a solution of the problem

−∆v = f (x ∈D), f := ∆w ∈L

(D),

|Σ

= 0.

)

(4.7.12)

According to Corollary 4.7.1, the (classical) solution of problem (4.7.12) exists and is

unique. Thus, we have proved the following assertion.

Theorem 4.7.6. Let ϕ be such that Q

0. Then a solution of the problem (4.7.11)

exists and is unique.

For more details on variational methods in connection with problems of mathematical

physics we refer the reader to textbooks like [7], [8] and [9].

Chapter

The

Cauchy-Kowalevsky Theorem

Deﬁnition 5.1. A function f (x

,. .. ,x

) is called analytic at the point (x

,. .. ,

), if in some neighbourhood of this point it can be represented in the form of an ab-

solutely convergent power series

f (x

,. .. ,x

) =

∞

∑

,...,k

−x

)

.. .(x

−x

)

. (5.1)

Consequently, the function f has partial derivatives of all orders and

,...,k

∂

+...+k

∂x

.∂x

,...,x

Moreover, the series (5.1) can be differentiated termwise in its domain of convergence an

arbitrary number of times.

Without loss of generality we will consider the case when

,. .. ,x

) = (0,.. ., 0).

Then (5.1) takes the form

f (x

,. .. ,x

) =

∞

∑

,...,k

.. .x

. (5.2)

Deﬁnition 5.2. A function F(x

,. .. ,x

) is called a majorant for a function

f (x

,. .. ,x

) of the form (5.2), if F(x

,. .. ,x

) is analytic in a neighbourhood of the origin:

F(x

,. .. ,x

) =

∞

∑

,...,k

.. .x

(5.3)

in a neighbourhood of 0 , and

,...,k

| ≤ B

,...,k

for all k

,. .. ,k

. Clearly, the radius of convergence of (5.2) is not less than the radius of

convergence of (5.3).

220 G.

Freiling and V. Yurko

Lemma 5.1. Let

the function f (t,x

,. .. ,x

) be analytic at the origin:

f (t,x

,. .. ,x

) =

∞

∑

,...,k

.. .x

. (5.4)

Then there exist M > 0 and a > 0 such that for each ﬁxed α ∈(0,1] the function

F(t,x

,. .. ,x

) =

1 −

t/α+x

..+x

(5.5)

majorant for f .

Proof. Since the series (5.4) converges absolutely in a neighbourhood of the origin,

there exist positive numbers a

,. .. ,a

such that

∞

∑

,...,k

.. .a

< ∞.

In particular, there exists M > 0 such that

,...,k

.. .a

≤ M

for all k

,. .. ,k

. Therefore,

,...,k

| ≤

Let a = min

0≤m≤n

. Then

,...,k

| ≤

+..

.+k

. (5

.6)

Consider a function F(t, x

,. .. ,x

) of the form (5.5) with these M > 0 and a > 0. Since

(5.5) is the sum of terms of the geometric progression, we have for |t|/α+|x

|+.. .+|x

F(t,x

,. .. ,x

) = M

∞

∑

k=0

(t/α + x

+ ... + x

)

= M

∞

∑

k=0

∑

..+k

Taking (5.6) into account we obtain

,...,k

+..

.+k

+ .

.. + k

≥

+..

.+k

≥

,...,k

and Lemma 5.1 is proved. 2

The general Cauchy-Kowalevsky theorem is a fundamental theorem on the existence

of the solution of the Cauchy problem for a wide class of systems of partial differential

equations. For simplicity we conﬁne ourselves here to the case of linear systems of the ﬁrst

order. We consider the following Cauchy problem:

∂u

∂t

∑

j=1

∑

k=1

(t, x

,. .. ,x

)

∂u

∂x

∑

j=1

(t, x

.. ., x

The Cauch

y-Kowalevsky Theorem 221

(t, x

.. ,x

), (5.7)

i|t=t

= ϕ

,. .. ,x

), i =

1,N, (5.8)

Here t,x

.. ,x

are independent variables and u

,. .. ,u

are unknown functions.

Theorem 5.1. Let the functions a

i jk

, b

i j

, c

be analytic at the point (t

,. .. ,

), and let the functions ϕ

be analytic at (x

,. .. ,x

). Then in some neighbourhood of

the point (t

,. .. ,x

) the Cauchy problem (5.7) −(5.8) has a unique analytic solution.

Proof. Without loss of generality we assume that t

= x

= . .. = x

= 0, i.e. we

seek a solution in a neighbourhood of the origin. Without loss of generality we also assume

that ϕ

,. .. ,x

) ≡ 0. Otherwise, for the functions v

:= u

−ϕ

we obtain a system of

the form (5.7) with different c

, but with zero initial conditions. Thus, we will solve the

Cauchy problem for system (5.7) in a neighbourhood of the origin with the initial conditions

(0,x

,. .. ,x

) = 0. (5.9)

Step 1: Constructing of the solution. Suppose that there exists an analytic (at the

origin) solution of the Cauchy problem (5.7), (5.9):

(t, x

,. .. ,x

) =

∞

∑

,...,k

.. .x

, (5.10)

,...,k

∂

+...+k

∂t

∂x

.∂x

|t=0,x

=...=x

Let us give a constructive procedure for ﬁnding the coefﬁcients α

,...,k

1) Differentiating (5.9) k

times with respect to x

, .. . , k

times with respect to x

and taking t = x

= . .. = x

= 0, we obtain α

0,k

...,k

= 0.

2) We differentiate (5.7) k

times with respect to x

, . .. , k

times with respect to x

and take t = x

= .. . = x

= 0. Then from the left we obtain α

1,k

...,k

, and from the right

we obtain known quantities. Thus, we have constructed the coefﬁcients α

1,k

,...,k

3) We differentiate (5.7) once with respect to t , k

times with respect to x

,. .. , k

times with respect to x

and take t = x

= . .. = x

= 0. Then from the left we obtain

2,k

...,k

, and from the right we obtain known quantities which depend on α

1,k

,...,k

. Thus,

we have constructed the coefﬁcients α

2,k

,...,k

4) We continue this process by induction. Suppose that the coefﬁcients α

j,k

,...,k

have been already constructed for j =

0,k

−1. W

differentiate (5.7) k

−1 times with

respect to t , k

times with respect to x

,. .. , k

times with respect to x

and take

t = x

= ... = x

= 0. Then from the left we obtain α

...,k

, and from the right we obtain

known quantities which depend on α

j,k

,...,k

, j =

0,k

−1. Thus,

have constructed the

coefﬁcients α

,...,k

This procedure of constructing the analytic solution (5.10) of the Cauchy problem (5.7),

(5.9) is called the A-procedure. In particular, this yields the uniqueness of the analytic so-

lution. It remains to show that the series (5.10), constructed by the A-procedure, converges

in some neighbourhood of the origin.